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Hydrogen atoms emit light in the Balmer transition from n = 4 to n =…
Question
Hydrogen atoms emit light in the Balmer transition from n = 4 to n = 2. This light falls on a metal surface of work function 1.0 eV. Using 13.6 eV for the hydrogen ground-state energy, find the maximum kinetic energy K (in eV) of the emitted photoelectrons, and report the integer value of 100*K.
✓ Verified answer: 155checked by our engine — not a guess
Step-by-step solution
Energy of the n=4 -> n=2 photon: E = 13.6*(1/4 - 1/16) = 13.6*(4/16 - 1/16) = 13.6*3/16 = 40.8/16 = 2.55 eV.
By Einstein's photoelectric equation, maximum kinetic energy K = E - phi = 2.55 - 1.0 = 1.55 eV.
Therefore 100*K = 155.
Final answer155
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